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- Thread starter MarkFL
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Sorry, but that is incorrect.

Also, I would prefer that work be shown, rather than simply relying on a CAS. The result should then be given in terms of $h_1,\,h_2,\,w$.

Whoops. Sorry. It should have been atan of course.

Also, I would prefer that work be shown, rather than simply relying on a CAS. The result should then be given in terms of $h_1,\,h_2,\,w$.

I do do the maths on paper first by the way (usually). It's just easier to post a link to WA than it is to post a proper answer.

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That was only one error...Whoops. Sorry. It should have been atan of course...

edit: Perhaps not though...it depends on how you have defined $a$ and $b$.

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- Feb 7, 2012

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If you draw a circle touching the ground and passing through the tops of the poles (at the points $P$ and $Q$) then it should be obvious geometrically that the wires should be attached at the point where the circle touches the ground.

Unfortunately, using this approach there does not seem to be an easy way to find the distance $x$ from the left pole to the point of attachment. The centre $C$ of the circle must lie on the perpendicular bisector of $PQ$, and the distance $CP$ must be equal to the height of $C$ above the ground. Using those facts, I found by some laborious calculations that $$x = \frac{\sqrt{h_1h_2\bigl(w^2 + (h_2-h_1)^2\bigr)} - h_1w}{h_2-h_1}.$$

Unfortunately, using this approach there does not seem to be an easy way to find the distance $x$ from the left pole to the point of attachment. The centre $C$ of the circle must lie on the perpendicular bisector of $PQ$, and the distance $CP$ must be equal to the height of $C$ above the ground. Using those facts, I found by some laborious calculations that $$x = \frac{\sqrt{h_1h_2\bigl(w^2 + (h_2-h_1)^2\bigr)} - h_1w}{h_2-h_1}.$$

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My solution:

We may state:

\(\displaystyle \alpha+\theta+\beta=\pi\)

\(\displaystyle \theta=\pi-(\alpha+\beta)\)

Now, we see that:

\(\displaystyle \tan(\alpha)=\frac{h_1}{x}\)

\(\displaystyle \tan(\beta)=\frac{h_2}{w-x}\)

Hence:

\(\displaystyle \theta=\pi-\left(\tan^{-1}\left(\frac{h_1}{x} \right)+\tan^{-1}\left(\frac{h_2}{w-x} \right) \right)\)

Differentiating with respect to $x$, we obtain:

\(\displaystyle \frac{d\theta}{dx}=-\frac{1}{1+\left(\frac{h_1}{x} \right)^2}\left(-\frac{h_1}{x^2} \right)-\frac{1}{1+\left(\frac{h_2}{w-x} \right)^2}\left(\frac{h_2}{(w-x)^2} \right)=\)

\(\displaystyle \frac{h_1}{x^2+h_1^2}-\frac{h_2}{(w-x)^2+h_2^2}=\frac{h_1\left((w-x)^2+h_2^2 \right)-h_2\left(x_2+h_1^2 \right)}{\left(x^2+h_1^2 \right)\left((w-x)^2+h_2^2 \right)}\)

Equating this to zero implies:

\(\displaystyle h_1\left((w-x)^2+h_2^2 \right)=h_2\left(x^2+h_1^2 \right)\)

Now, expanding and arranging in standard quadratic form, we obtain:

\(\displaystyle \left(h_2-h_1 \right)x^2+\left(2h_1w \right)x+h_1\left(h_1h_2-h_2^2-w^2 \right)=0\)

We find that the discriminant $\Delta$ is:

\(\displaystyle \Delta=\left(2h_1w \right)^2-4\left(h_2-h_1 \right)\left(h_1\left(h_1h_2-h_2^2-w^2 \right) \right)\)

After simplification we find:

\(\displaystyle \Delta=4h_1h_2\left(\left(h_2-h_1 \right)^2+w^2 \right)\)

and so, application of the quadratic formula on the quadratic in $x$, and discarding the negative root, there results:

\(\displaystyle x=\frac{-h_1w+\sqrt{h_1h_2\left(\left(h_2-h_1 \right)^2+w^2 \right)}}{h_2-h_1}\)

To find what portion of the distance between the two poles, as measured from the first pole, we must take as the staking point for the two wires, we may use:

\(\displaystyle \frac{x}{w}=\frac{-h_1w+\sqrt{h_1h_2\left(\left(h_2-h_1 \right)^2+w^2 \right)}}{w\left(h_2-h_1 \right)}\)